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\author{五六七 }
\title{斐波那契数列 }
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%\date{2020 年 2 月 28 日}

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\begin{document}

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\begin{abstract}
根据数列的递推关系式计算数列的通项。
\end{abstract}

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\section{问题描述}

设有一对兔子出生一个月后开始繁殖，每个月出生一对小兔子。小兔子一个月后也开始繁殖，每个月又出生一对小兔子。假设兔子只有繁殖，没有死亡，求第 $k$ 个月的时候一共有多少对兔子？

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\section{建立模型}

设第 $k$ 个月开始的时候，一共有 $F_k$ 对兔子。则根据题目假设，有
\begin{table}[ht]\centering
\caption{前几个月的兔子数量（单位：对）}\vspace{0.2cm}
\begin{tabular}{|M{2cm}|M{1cm}|M{1cm}|M{1cm}|M{1cm}|M{1cm}|M{1cm}|M{1cm}|}\hline 
月份 &0&1&2&3&4&5 & $\cdots$\\ \hline 
兔子数量 &1&1&2&3&5&8& $\cdots$ \\ \hline 
\end{tabular}
\end{table}

写成递推关系式，可得
\begin{eqnarray}
F_0=1, F_1=1, F_{k+2}=F_{k+1}+F_k, k=0,1,2,\cdots.  
\end{eqnarray}

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\section{求解模型}

\subsection{矩阵的对角化方法}
数列的递推关系式可以写成如下形式
\begin{eqnarray}
\begin{bmatrix} F_{k+1} \\ F_{k+2} \end{bmatrix} 
=
\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} 
\begin{bmatrix} F_k \\ F_{k+1} \end{bmatrix}.  
\end{eqnarray}
由此可得
\begin{eqnarray}
\begin{bmatrix} F_{k+1} \\ F_{k+2} \end{bmatrix} 
=
\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix} 
\begin{bmatrix} F_k \\ F_{k+1} \end{bmatrix}
=
\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}^2
\begin{bmatrix} F_{k-1} \\ F_k \end{bmatrix}
= \cdots =
\begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}^{k+1}
\begin{bmatrix} F_{0} \\ F_1 \end{bmatrix}. 
\end{eqnarray}

因此问题转化成计算矩阵 $A$ 的 $n$ 次幂，其中 
\begin{eqnarray}
A = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}.  
\end{eqnarray}
计算矩阵 $A$ 的特征值和特征向量，可得 特征值为 
\begin{eqnarray}
\lambda_1 = \frac{1-\sqrt{5}}{2}, \,\,\, \lambda_2 = \frac{1+\sqrt{5}}{2}.
\end{eqnarray}
相应的特征向量为
\begin{eqnarray}
\xi_1 = \begin{bmatrix} \sqrt{5}+1 \\ -2 \end{bmatrix}, \,\,\, 
\xi_2 = \begin{bmatrix} \sqrt{5}-1 \\ 2 \end{bmatrix}. 
\end{eqnarray}
设矩阵 
\begin{eqnarray}
P = \begin{bmatrix} \xi_1 & \xi_2 \end{bmatrix}
= \begin{bmatrix} \sqrt{5}+1 & \sqrt{5}-1 \\ -2 & 2 \end{bmatrix}, \,\,\,
B = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}
= \begin{bmatrix} \frac{1-\sqrt{5}}{2} & 0 \\ 0 & \frac{1+\sqrt{5}}{2} \end{bmatrix}. 
\end{eqnarray}
则有 
\begin{eqnarray}
P^{-1}AP = B.
\end{eqnarray}
由此可得 $A = PBP^{-1}$, 以及 
\begin{eqnarray}
A^n = (PBP^{-1})^n = (PBP^{-1})(PBP^{-1})\cdots (PBP^{-1}) = PB^nP^{-1}. 
\end{eqnarray}

\subsection{差分方程降阶方法}

设从递推关系式
\begin{eqnarray}
F_{k+2}=F_{k+1}+F_k 
\end{eqnarray}
可以得到 
\begin{eqnarray}
F_{k+2} - a F_{k+1} =b(F_{k+1} - aF_k),   
\end{eqnarray}
其中 $a,b$ 待定。比较两式可得
\begin{eqnarray}
a+b=1,\,\, ab=-1.
\end{eqnarray}
由此求出 $a,b$. 

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\section{回答问题}
通过矩阵的对角化方法，可以求得数列的通项，也就是第 $k$ 个月的兔子数量为
\begin{eqnarray}
F_k = \left(\frac{1}{2} -\frac{\sqrt{5}}{10} \right)  \left( \frac{1-\sqrt{5}}{2} \right)^k 
+  \left( \frac{1}{2} +\frac{\sqrt{5}}{10} \right)  \left( \frac{1+\sqrt{5}}{2} \right)^k. 
\end{eqnarray}

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\section{附加问题}

\begin{enumerate}%\itemsep0.5em 

\item  求差分方程 $x_{n+2}-x_{n+1}-2x_n=0, x_0=x_1=-2$ 的解。用不同方法求出通项，并编程验证。

\item  设 $F_n$ 为第 $n$ 个斐波那契数，设 $S=\sum\limits_{n=1}^{\infty} \frac{1}{F_n^2}$. 
\begin{enumerate}
\item  用 $S$ 表示 $\sum\limits_{n=1}^{\infty} \frac{1}{(F_{n+2}F_{n+1}F_n)^2}$. 
\item  用 $S$ 表示 $\sum\limits_{n=1}^{\infty} \frac{1}{(F_{n+2}F_n)^2}$. 
\end{enumerate}


\end{enumerate}

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%\section{参考文献 }
\begin{thebibliography}{99}
\bibitem{sishoukui-2} 司守奎,孙玺菁. \emph{Python数学建模算法与应用}, 国防工业出版社. 2022年1月第1版. 

\end{thebibliography}

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